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3 years ago

Please help me with this

Mathematics

2 answers:

juin [17]3 years ago

8 0

Answer:

x=30, <A=60, <C=30

Step-by-step explanation:

A triangle always has its total angles added up to 180 degrees, so we can set up the following equation:

90+2x+x=180

90+3x=180

3x=90

x=30

Since x=30, this means <C is 30 degrees. For <A, it's twice as much as <C, so 2*30=60.

So x=30, <A=60, <C=30

Hope this helped!

Gnom [1K]3 years ago

4 0

Answer:

can u pls make it bigger

Step-by-step explanation:

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Nicole has 18 nickels n and dimes d. If the value of her coins is $1.45, how many of each coin does she

Sloan [31]

Answer:

19 nickel = .95

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Step-by-step explanation:

.95 + .50 = 1.45

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3 years ago

when full the gas tank of a car holds 15 gallons. it now contains 12 gallons. what percent represents how full the tank is?

Masteriza [31]

Hello!

to find the percent of a whole, use this equation here:

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3 years ago

Write an equation for a like that is parallel to the given point. Explain how you found the equation.

pishuonlain [190]

Answer:

The line would be y = 2x + 5

Step-by-step explanation:

To find a parallel line, we first need to note that the slope of the original line is 2. This means the slope of our new line will also be 2 because parallel lines have the same slope.

So we use the slope we found along with the point given in point-slope form. Then we solve for y.

y - y1 = m(x - x1)

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7 0

3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

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$x^{2}+8x-7x-56=0$

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$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

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Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

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d1i1m1o1n [39]

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Hence the point lies within the circle .

4 0

2 years ago

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